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Theorem sbbid 2078
 Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 5-Aug-1993.)
Hypotheses
Ref Expression
sbbid.1 xφ
sbbid.2 (φ → (ψχ))
Assertion
Ref Expression
sbbid (φ → ([y / x]ψ ↔ [y / x]χ))

Proof of Theorem sbbid
StepHypRef Expression
1 sbbid.1 . . 3 xφ
2 sbbid.2 . . 3 (φ → (ψχ))
31, 2alrimi 1765 . 2 (φx(ψχ))
4 spsbbi 2077 . 2 (x(ψχ) → ([y / x]ψ ↔ [y / x]χ))
53, 4syl 15 1 (φ → ([y / x]ψ ↔ [y / x]χ))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176  ∀wal 1540  Ⅎwnf 1544  [wsb 1648 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925 This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649 This theorem is referenced by:  sbcom  2089  sbcom2  2114
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